Replacing values of a dataframe column with a vector

Question

I want to simplify my current method of replacing column values of a dataframe with a vector. I've provided a reproducible answer below with my solution using base R. A simplified version contains just one data frame and with multiple dataframes, i'm forced to use a for loop due to my bad solution.

How can I simplify my approach?

# Simplified version
Df <- data.frame(a = c(1,2,3),
           b = c(4,5,6),
           c = c(7,8,9))

l <- list(c(11,22,33),
          c(44,55,66))

letters <- c("a","b")

Df[letters] <- l


# Multiple data frames
Df1 <- list(data.frame(a = c(1,2,3),
                       b = c(4,5,6),
                       c = c(7,8,9)),
            data.frame(a = c(101,102,103),
                       b = c(104,105,106),
                       c = c(107,108,109)))
l <- list( list(c(11,22,33), c(44,55,66)),
           list(c(111,222,333), c(444,555,666)) )

letters <- c("a","b")

for(i in 1:length(Df1)){

  Df1[[i]][letters] <- l[[i]]

}

akrun · Accepted Answer · 2019-11-12 20:19:34Z

2

Here is an option with map2

library(purrr)
library(dplyr)
map2(Df1, l, ~ {.x[letters] <- .y; .x})

Or with inset from magrittr

library(magrittr)
map2(Df1, l, ~  inset(.x, letters, value = .y))

or in a chain

map2(Df1, l, ~ .x %>% 
                 select(-one_of(letters)) %>% 
                 bind_cols(.y %>% 
                               set_names(letters)) %>% 
                 select(names(.x)))

Or in base R

Map(function(x, y) {x[letters] <- y;x},  Df1, l)

edited Nov 12, 2019 at 20:19

answered Nov 12, 2019 at 17:45

akrun

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8 Comments

Ali Over a year ago

Thank you for your answer. This seems very complicated when compared to my for loop. For readability, would your solution be considered a better approach/easier to read for an experienced R user?

akrun Over a year ago

@Ali You asked for a solution with dplyr. If it is okay for a base R, then I can change to more simpler ways

Ali Over a year ago

I think that would be more appropriate as I assumed using dplyr would simplify this. I’ll change my question accordingly

akrun Over a year ago

@Ali Thanks. I changed the code. Now, it is one-liner, but I used the earlier code so that it would be in a chain instead of wrapping within {}

Ali Over a year ago

much much appreciated for this advice. I’ll take this all into account and hopefully next time will be a more complex question for you :) thank you.

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